After you multiply $\frac{7}{3}$ by $6x$ and by $-9$, rewrite the equation and then combine all the $x$-terms together and all the constant terms together.

Once you simplify, ask: does the equation end up with an $x$ term, no $x$ term but a true statement, or no $x$ term and a false statement? That tells you how many solutions there are.

In Desmos, type y = (7/3)(6x - 9) - 14x + 21 to graph the left side of the equation as a function of $x$.

Look at the graph that appears. Notice where this graph lies relative to the $x$-axis (the line $y = 0$).

The solutions to the equation are the $x$-values where the graph has $y = 0$ (where it is on the $x$-axis). Decide whether that happens for no $x$-values, only one $x$-value, exactly two $x$-values, or for all $x$-values.

Start by distributing $\frac{7}{3}$ to both terms inside the parentheses:

Compute each product:

• $\frac{7}{3} \cdot 6x = 14x$ (because $6 \div 3 = 2$ and $7 \cdot 2x = 14x$)
• $\frac{7}{3} \cdot (-9) = -21$ (because $9 \div 3 = 3$ and $7 \cdot (-3) = -21$)

So the equation becomes:

Group the $x$-terms and the constant terms:

• $14x - 14x = 0$
• $-21 + 21 = 0$

So the left side simplifies to:

which is just

The simplified equation $0 = 0$ does not contain $x$ at all and is always true, no matter what real number $x$ is.

That means every real number is a solution, so the equation has infinitely many real solutions.